2. STRESS, STRAIN, AND CONSTITUTIVE RELATIONS

Mechanics of materials is a branch of mechanics that develops relationships

between the external loads applied to a deformable body and the intensity of internal

forces acting within the body as well as the deformations of the body. External forces can

be classified as two types: 1) surface forces produced by a) direct contact between two

bodies such as concentrated forces or distributed forces and/or b) body forces which

occur when no physical contact exists between two bodies (e.g., magnetic forces,

gravitational forces, etc.). Support reactions are external surface forces that develop at

the support or points of support between two bodies. Support reactions may include

normal forces and couple moments. Equations of equilibrium (i.e., statics) are

mathematical expressions of vector relations showing that for a body not to translate ormove along a path then F = 0∑ . For a body not to rotate, M = 0∑ . Alternatively, scalar

equations in three-dimensional space (i.e., x, y, z) are:

F x = 0∑ F y = 0 ∑ F z = 0∑Mx = 0∑ My = 0∑ Mz = 0∑

(2.1)

Internal forces are non external forces acting in a body to resist external loadings.

The distribution of these internal forces acting over a sectioned area of the body (i.e., force

divided by area, that is, stress) is a major focus of mechanics of materials. The response

of the body to stress in the form of deformation or normalized deformation, that is, strain is

also a focus of mechanics of materials. Equations that relate stress and strain are known

as constitutive relations and are essential, for example, for describing stress for a

measured strain.

Stress

If an internal sectioned area is subdivided into smaller and smaller areas, ∆A , two

important assumptions must be made regarding the material: it is continuous and it is

cohesive. Thus, as the subdivided area is reduced to infinitesimal size, the distribution of

internal forces acting over the entire sectioned area will consist of an infinite number of

forces each acting on an element, ∆A , as a very small force ∆F . The ratio of incremental

force to incremental area on which the force acts such that: lim∆A→0

∆F

∆A is the stress which

can be further defined as the intensity of the internal force on a specific plane (area)

passing through a point.

2.1

Stress has two components, one acting perpendicular to the plane of the area and

the other acting parallel to the area. Mathematically, the former component is expressed

as a normal stress which is the intensity of the internal force acting normal to an

incremental area such that:

σ = lim∆A→0

∆Fn

∆A (2.2)

where +σ = tensile stress = "pulling" stress and -σ = compressive stress = "pushing"

stress. The latter component is expressed as a shear stress which is the intensity of the

internal force acting tangent to an incremental area such that:

τ = lim∆A→0

∆Ft

∆A (2.3)

The general state of stress is one which includes all the internal stresses acting on

an incremental element as shown in Figure 2.1. In particular, the most general state of

stress must include normal stresses in each of the three Cartesian axes, and six

corresponding shear stresses.

Note for the general state of stress that +σ acts normal to a positive face in the

positive coordinate direction and a +τ acts tangent to a positive face in a positivecoordinate direction. For example, σ xx (or just σ x ) acts normal to the positive x face inthe positive x direction and τ xy acts tangent to the positive x face in the positive y

direction.

Although in the general stress state, there are three normal stress component and

six shear stress components, by summing forces and summing moments it can be shownthat τ xy = τyx ;τxz = τzx ;τyz = τzy .

z

x

y

ττ

τ

τ

zy

zx

yxxy

τ

τ xz

yz

σ

σ

σ

y

z

x

Figure 2.1 General and complete stress state shown on a three-dimensional incrementalelement.

2.2

Therefore the complete state of stress contains six independent stress components (threenormal stresses, σ x ;σy ;σ z and three shear stresses, τ xy;τyz;τxz ) which uniquely

describe the stress state for each particular orientation. This complete state of stress can

be written either in vector formσx

σy

σz

τ xy

τ xz

τ yz

(2.4)

or in matrix form

σ x τxy τxz

τxy σ y τyz

τxz τ yz σ z

(2.5)

The units of stress are in general: Force

Area=

F

L2 . In SI units, stress is

Pa =N

m2 or MPa = 106 N

m2 =N

mm2 and in US Customary units, stress is

psi =lbf

in 2 or ksi = 103 lbf

in 2 =kip

in 2 .

Often it is necessary to find the stresses in a particular direction rather than just

calculating them from the geometry of simple parts. For the one-dimensional case shown

in Fig. 2.2, the applied force, P, can be written in terms of its normal, PN, and tangential,PT, components which are functions of the angle, θ, such that:

PN = P cos θPT = P sinθ

(2.6)

The area, Aθ, on which PN and PT act can also be written in terms of the area, A, normal tothe applied load, P, and the angle, θ, such that

Aθ = A /cosθ (2.7)

The normal and shear stress relation acting on any area oriented at angle, θ, relative to

the original applied force, P are:

σθ =PN

Aθ=

P cos θA / cosθ

=P

Acos2 θ = σ cos2 θ

τθ = PT

Aθ= P sinθ

A /cosθ= P

Acosθ sinθ = σ cosθ sinθ

(2.8)

where σ is the applied unidirectional normal stress.

2.3

θ

θP

P

P

A

A = A cos

θ

θ θ

N

T

Figure 2.2 Unidirectional stress with force and area as functions of angle, θ

For the two dimensional case (i.e., plane stress case such as the stress state at a

surface where no force is supported on the surface), stresses exist only in the plane of thesurface (e.g., σx ;σy ; τxy ). The plane stress state at a point is uniquely represented by

three components acting on a element that has a specific orientation (e.g., x, y) at the

point. The stress transformation relation for any other orientation (e.g., x', y') is found byapplying equilibrium equations ( F = 0 and∑ M = 0 )∑ keeping in mind that

F n = σA and F t = τA. The rotated axes and functions for incremental area are shown in

Fig. 2.3. The forces in the x and y directions due to F n = σA and F t = τA and acting on the

areas normal to the x and y directions are shown in Fig. 2.4By applying simple statics such that in the x'-direction, F x ' = 0∑ and

σx ' = σ x cos2θ + σy sin2 θ + 2τxy cosθ sinθ

or

σx ' =σ x + σy

2+

σx − σy

2cos2θ + τxy sin2θ

(2.9)

X

y

x'

y'θ

θ

∆A

∆Ay=∆A sin θ

∆Ax=∆A cos θ

Figure 2.3 Rotated axes and functions for incremental area.

2.4

X

y x'

y' θ

θ

σy ∆Ay

σx ∆Ax

τxy ∆Ay

σx' ∆A

τx'y' ∆A

θ

θ

θ

θ

τxy∆Ax

Figure 2.4 Rotated coordinate axes and components of stresses/forces.

Similarly, for the x'y'-direction, Fy ' = 0∑ and

τx 'y ' = (σ x − σy )cos θ sinθ + τxy (cos2 θ + sin2θ)

or

τx ' y ' = −σx − σy

2sin2θ + τ xy cos2θ

(2.10)

Finally, for the y' direction, Fy ' = 0∑ and

σy ' = σ x sin2θ + σy cos2 θ − 2τ xy cosθ sinθ

or

σy ' =σ x +σ y

2−

σx − σy

2cos2θ − τxy sin2θ

(2.11)

If the stress in a body is a function of the angle of rotation relative to a given

direction, it is natural to look for the angle of rotation in which the normal stress is either

maximum or nonexistent. A principal normal stress is a maximum or minimum normal

stress acting in principal directions on principal planes on which no shear stresses act.

Because there are three orthogonal directions in a three-dimensional stress state thereare always three principal normal stresses which are ordered such that σ1 > σ2 > σ3 .

Mathematically, the principal normal stresses are found by determining the angulardirection, θ, in which the function, σ = f (θ ), is a maximum or minimum by differentiating

σ = f (θ ) with respect to θ and setting the resulting equation equal to zero such that dσdθ

= 0

before solving for the θ at which the principal stresses occur. Applying this idea to Eq. 2.9,

gives

2.5

dσdθ

= 0 = σx −σ y( )sin2θ + 2τxy cos2θ ⇒ sin2θcos2θ

= tan2θ =2τxy

σx − σy(2.12)

There are two solutions for the principal stress angle (i.e., for maximum and minimum) so

that.

θN1 =12

tan-1 2τxy

σx − σy

θN2 =12

tan-1 2τxy

σx − σy+ π

= θN1 + π

2

(2.12)

Using trigonometry on the geometry shown in Fig. 2.5 results in

tan2θ =τ xy

σx −σ y( ) /2

sin2θ =τ xy

σ x − σy

2

2

+ τ xy2

cos2θ =σx − σy( ) /2

σx − σy

2

2

+ τ xy

2

(2.14)

Substituting the trigonometric relations of Eq. 2.14 back into Eq. 2.9 gives for the plane

stress case:

σθ = σ1,2 =σx +σ y

2±

σ x − σy

2

2

+τ xy2

tan2θp =τxy

σx − σ y( ) /2

(2.15)

2θ

τ

(σ − σ )/2

√ x

y

xy

σ − σ ( )22 xy2+ τy

x

Figure 2.5 Geometric representation of the principal direction relations

2.6

Note that for the plane stress case in the x-y plane, σz = 0 . Thus the 1 and 2 subscripts in

Eq. 2.15 are only for the x-y plane and are not necessarily σ1 and σ2 for the three-

dimensional general state of stress. Therefore, ordering of σ1 and σ2 of Eq. 2.15 is only

preliminary, until they are compared to σz and ordered according to conventionσ1 > σ2 > σ3 . For example, for a particular plane stress state σ1 and σ2 found from Eq.

2.15 are 100 and 20 MPa, then the principal stresses are σ1 = 100, σ2 = 20, σ3 = 0 MPa.

However, if σ1 and σ2 found from Eq. 2.15 are 125 and -5 MPa, then the principalstresses are σ1 = 125, σ2 = 0, σ3 = −5 MPa. Finally, if σ1 and σ2 found from Eq. 2.14 are

-25 and -85 MPa, then the principal stresses are σ1 = 0, σ 2 = -25, σ 3 = −85 MPa.

Performing a similar substitution of the trigonometric relations of Eq. 2.14 back into

Eq. 2.10 gives for the plane stress case:

τmax =σ x −σ y

2

2

+τ xy2

σave =σ x +σ y

2 and tan2θs =

− σx − σy( )2τxy

(2.16)

Note that the τ max of Eq. 2.15 is only for the x-y plane. The maximum shear stress for thethree-dimensional stress state can be found after the principal stresses are orderedσ1 > σ2 > σ3 such that:

τ1,3 =σ1 − σ3

2 (2.17)

Some general observations can be made about principal stresses.

a) In a principal direction, when τ =0, then σ 's are maximum or minimumb) σ max and σmin τmax and τmin( ) occur in directions 90° apart.

c) τ max occurs in a direction midway between the directions of σ max and σmin

σ

τ +σ ,+τ

+σ ,−τ

φ =2θ

x

y

C= 2

x + yσσR = ( x - C) + )

tan = -

σ τ 22

φ −τ σ ( x - C)

2

Figure 2.6 Mohr's circle representation of plane stress state

2.7

An interesting graphical relation occurs if the second equation in each of Eqs. 2.9

and 2.10 are squared and added together:

σ x ' = σθ =σx +σ y

2+

σ x − σy

2cos2θ + τ xy sin2θ

2

+

τ x 'y '=τθ =−σx −σy

2sin2θ +τ xy cos 2θ

2

=

σθ −σ x +σ y

2

2

+τθ2 =

σx −σy

2

2

+τ xy2

≡

x−h( )2 +y2 = r2

(2.18)

The result shown in Eq. 2.18 is the equation for a circle (i.e., Mohr's circle) with radius,

r=σ x −σy

2

2

+ τxy2 and displaced h=

σ x +σ y

2 on the x=σ θ axis as illustrated in Fig.

2.6. Examples of Mohr's circles are shown in Fig. 2.7. A procedure for developing Mohr's

circle for plane stress is shown in the following section.

σ

τ

1σσ

τ

2

maxfor x-y plane

Mohr's circle for stresses in x-y plane

σ

τ

1σσ

τ

3

max =

Mohr's circle for stresses in x-y-z planes

σ2

σ − σ2

1 3

Figure 2.7 Examples of Mohr's circle for plane and three dimensional stress states.

2.8

Graphical Description of State ofStress

2-D Mohr's Circle

σ

σ

τxy

y

x

X

Y

In this example all stresses acting in axialdirections are positive as shown in Fig. M1.

Fig. M1- Positive stresses acting on a physical element.

+τ

+σx

y

xy

+σ

y-face As shown in Figs. M2 and M3, plottingactual sign of the shear stress with xnormal stress requires plotting of theopposite sign of the shear stress with the ynormal stress on the Mohr's circle.

Fig. M2 - Directionality of shear acting on x and y faces. In this example σx > σy and τxy is positive.

By the convention of Figs. M2 and M3, φ =2θ on the Mohr's circle is negative from the+σ axis. (Mathematical convention is thatpositive angle is counterclockwise).

σ

τ +σ ,+τ

+σ ,−τ

φ =2θ

x

y

C= 2

x + yσσR = ( x - C) + )

tan = -

σ τ 22

φ −τ σ ( x - C)

2

Note that by the simple geometry of Fig.M3, φ = 2θ appears to be negative while bythe formula, tan 2θ = 2τxy/(σx -σy), thephysical angle, θ, is actually positive.

In-plane principal stresses are: σ1= C+R σ2= C - R

Fig. M3 - Plotting stress values on Mohr's circle.

Maximum in-plane shear stress is:τmax=R=(σ1-σ2)/2

2.9

θ

σσ

1

2Y

X

Directionof +θ The direction of physical angle, θ, is from

the x-y axes to the principal axes.

Fig. M4 - Orientation of physical element with only principal stresses acting on it.

θ

PrincipalAxis

Line of X-Y Stresses

Direction of

Fig. M5 - Direction of q from the line of x-y stresses to the principal stress

Note that the sense (direction) of thephysical angle, θ, is the same as on theMohr's circle from the line of the x-ystresses to the axes of the principalstresses.

axis.

Same relations apply for Mohr's circle for strain except interchange variables as

σ ⇔ ε and τ ⇔γ2

2.10

Recall that all stress states are three-dimensional. Therefore, a more general

method is required to solve for the principal stresses. One such method is to solve for the

"eigenvalues" of the stress matrix where, σ is the principal stress:

σx −σ τ xy τxz

τxy σy −σ τ yz

τ xz τyz σz −σ

(2.19)

Finding the determinant for this matrix and grouping terms gives:

σ 3 − I1σ2 + I2σ − I3 = 0 (2.20)

where the stress invariants, I1 ,I2, I3( ), do not vary with stress direction:

I1 = σ x + σy + σz

I2 = σxσ y +σ yσz + σ zσx − τxy2 − τxz

2 −τ yz2

I3 = σ xσyσ z + 2τ xyτ xzτ yz − σxτ yz2 − σyτxz

2 −σ zτxy2

(2.21)

Note that if principal stresses are used in Eq. 2.21 for the σ terms then all terms

containing τ will be zero since by definition, principal stresses act in principal directions

on principal planes on which τ =0.Eq. 2.20 can then be solved for the three roots which when ordered σ1 > σ2 > σ3

are the principal stresses. Eq. 2.20 can be plotted as f σ( ) vs σ shown in Fig. 2.9 where

the principal stresses are the values of σ which occur when f σ( ) = 0.

σ

f( )σ

σσσ123

Figure 2.9 Solving for cubic routes for principal stresses

2.11

Strain

Whenever a force is applied to a body, it will tend to change the body's shape and

size. There changes are referred to as deformation. Size changes are known as

dilatation (volumetric changes) and are due to normal stresses. Shape changes are

known as distortion and are due to shear stresses.

In order to describe the deformation in length of line segments and the changes in

angles between them, the concept of strain is used. Therefore, strain is defined as

normalized deformations within a body exclusive of rigid body displacements

There are two type of strain, one producing size changes by elongation or

contraction and the other producing shape changes by angular distortion.

Normal strain is the elongation or contraction of a line segment per unit length

resulting in a volume change such that

ε = limB →A along n

A' B'− AB

AB≡

L f − Lo

Lo

(2.22)

where + ε = tensile strain = elongation and -ε = compressive strain = contraction

Shear strain is the angle change between two line segments resulting in a shape

change such that:

γ = (θ =π2

) − θ ' ≈∆h

(for small angles ) (2.23)

where +γ occurs if π2

> θ ' and -γ occurs if π2

< θ '.

The general state is one which includes all the internal strains acting on an

incremental element as shown in Fig. 2.10. The complete state of strain has sixindependent strain components (three normal strains, εx; εy; εz and three engineering

shear strains, γ xy;γ yz;γ xz ) which uniquely describe the strain state for each particular

orientation. This complete state of strain can be written in vector formεx

εy

εz

γ xy

γ xz

γ yz

(2.24)

2.12

A

ε xy ε

ε yx

ε

x

y

Engineering shear strain,

γ = ε + εxy xy yx

Figure 2.10 General and complete strain state shown on a three-dimensional incrementalelement.

Alternatively, the complete state of strain can be written in matrix form :

εx γ xy γ xz

γ xy εy γ yz

γ xz γ yz εz

(2.25)

The units of strain In general: Length

Length=

L

L, In SI units, strain is

m

m for ε and

m

m or radian for γ and in US Customary units, strain is

in

in for ε and

in

in or radian for γ

Just as in the stress case it is often necessary to find the stresses in a particular

direction rather than just calculating them from the geometry of simple parts. For the two-

dimensional, plane strain condition (e.g., strain at a surface where no deformation occursnormal to the surface), strains exist only in the plane of the surface (e.g., εx; εy;γ xy ). The

plane strain state at a point is uniquely represented by three components acting on a

element that has a specific orientation (e.g., x, y) at the point. The strain transformation

relation for any other orientation (e.g., x', y') is found by summing displacements in theappropriate directions keeping in mind that δ = ε Lo and ∆ = γ h as shown in Fig. 2.11

Simply adding components of displacements in the x' direction,displacements in x 'direction for Q to Q *∑ (see Fig. 2.12) gives

εx ' = εx cos2 θ + ε y sin2 θ + γ xy cosθ sinθ

or

εx ' =εx + εy

2+

εx − εy

2cos2θ +

γ xy

2sin2θ

(2.25)

2.13

θ

x

y

}

}

}

y'

x'

Q

Q*

dy

δ = ε y y

δ = ε x x

∆ = γ dy

dx

ds

dy

dx

Figure 2.11 Rotated coordinate axes and displacements for x and y directions

Similarly, adding components of displacements in due to rotations of dx' and dy'rotation of dx ' and dy'∑ (see Fig. 2.12) gives

γ x 'y '

2= (εx − εy )cos θ sinθ +

γ xy

2(cos2 θ + sin2θ)

or

γ x 'y '

2= −

εx − εy

2sin2θ +

γ xy

2cos2θ

(2.26)

Finally, adding components of displacements in the y' direction,displacements in y 'direction for Q to Q *∑ (see Fig. 2.12) gives

εy ' = εx sin2θ + εy cos2 θ − γ xy cosθ sinθ

or

εy ' =εx + εy

2−

εx − εy

2cos2θ −

γ xy

2sin2θ

(2.27)

Q

Q*

x'y

xδ = ε x x

δ = ε y y

∆ = γ

θ

θ

θ

{δ = ε x' x'

dx

dy

ds

dy

cos

sin

θ =

θ =

ds

ds

dx

dy

Figure 2.12 Displacements in the x' direction for strains/displacements in the x and ydirections

2.14

Just as for the stress in a body which is a function of the angle of rotation relative to

a given direction, it is natural to look for the angle of rotation in which the normal strain is

either maximum or minimum. A principal normal strain is a maximum or minimum normal

strain acting in principal directions on principal planes on which no shear strains act.

Because there are three orthogonal directions in a three-dimensional strain state thereare always three principal normal strains which are ordered such that ε1 > ε2 > ε3 .

Also as in the stress case, the principal strains for the plane strain case

ε1,2 =εx + εy

2±

εx − εy

2

2

+γ xy

2

2

tan2θp =γ xy

εx − εy

(2.29)

Note that for the plane strain case in the x-y plane, εz = 0 . Thus, the 1 and 2 subscripts in

Eq. 2.29 are only for the x-y plane and are not necessarily ε1 and ε2 for the three-

dimensional general state of strain. Therefore, ordering of ε1 and ε2 of Eq. 2.29 is only

preliminary, until they are compared to εz and ordered according to conventionε1 > ε2 > ε3 .

For the shear strain case

γmax2

=εx − εy

2

2

+γ xy

2

τxy

2 ,

εave =εx + εy

2 and tan2θs =

− εx − εy( )γ xy

(2.30)

σ

γ/2

1εε

γ

2

max/2 for x-y plane

Mohr's circle for strains in x-y plane

σ

γ/2

1εε

γ

3

max =

Mohr's circle for strains in x-y-z planes

ε2

ε − ε1 3

Figure 2.13 Examples of Mohr's circle for strain.

2.15

45°

60°

60°a

bc

a

bc

45° Rectangular 60° Deltax

y

x

Figure. 2.14 Rectangular and Delta rosettes.

Note that the γ max of Eq. 2.30 is only the x-y plane. The maximum shear strain for thethree-dimensional strain state can be found after the principal strains are orderedε1 > ε2 > ε3 such that:

γ 1,3 = ε1 − ε3 (2.31)

As with stress, the complete strain state can be represented graphically as a Mohr's

circle. Examples of Mohr's circles for strain are shown in Fig. 2.13. Note that the same

procedure for developing Mohr's circle for the plain strain case can be used as with the

plane stress case with by making the following substitutions: σ ⇔ ε and τ ⇔γ2

.

An important application of the strain transformation relation of Eq. 2.26 is to

experimentally determine the complete strain state since stress is an abstract engineering

quantity and strain is measurable/observable engineering quantity. Equation 2.26

requires that three strain gages be applied at arbitrary orientations at the same point on

the body to determine the principal strains and orientations. However, to simplify the data

reduction, the three strain gages are applied at fixed angles usually in the form of 45°

rectangular rosette or 60° Delta rosette as shown in Fig. 2.14. The resulting equations to

determine the local (for the strain gage rosettes shown in Fig. 2.14) x-y strains in

preparation for determining the principal strains:

45° Rectangular 60° Delta

Rosette Rosette

εx = εa εx = εa

εy = εc εy =13

(2εb + 2εc − εa )

γ xy = 2εb − (εa + εc ) γ xy =23

(εb − εc )

(2.32)

However, a limitation of measuring strains experimentally is that stresses are often

required to determine the engineering performance of the component. Thus, equations

which relate stress and strain are required.

2.16

Constitutive Relations

If the deformation and strain are the response of the body to an applied force or

stress, then there must be some type of relations which allow the strain to be predicted

from stress or vice versa. The uniaxial stress-strain case is a useful example to begin to

understand this relation.

During uniaxial loading (see Fig. 2.15) by load, P, of a rod with uniform cross

sectional area, A, and length, L, the applied stress, σ ,is calculated simply as

σ = P

A(2.33)

The resulting normal strain, εL , in the longitudinal direction can be calculated from

the deformation response, ∆L , along the L direction:

εL = ∆L

L(2.34).

Another normal strain, εT , in the transverse direction can be calculated from the

deformation response, ∆D , along the D direction:

εT = ∆D

D(2.35).

A plot of σ vs. εL (see Fig. 2.16) shows a constant of proportionality between stress

and strain in the elastic region such that

y = mx + b ⇒ σ = EεL (2.36)

where E =∆σ∆εL

is known as the elastic modulus or Young's modulus and Eq. 2.36 is

known as unidirectional Hooke's law.

A

PDfD=Do

L=Lo

Lf

L=Lf - L

D=Df -D

∆

∆

P Deformed

Undeformed

Figure 2.15 Uniaxially-loaded rod undergoing longitudinal and transverse deformation

2.17

Longitudinal strain, L Longitudinal strain, εε L

Ε−ν

Figure 2.16 Plots of applied stress vs. longitudinal strain and transverse strain versuslongitudinal strain

A plot of εT vs. εL (see Fig. 2.16) shows a constant of proportionality between

transverse and longitudinal strains in the elastic region such that

y = mx + b ⇒ εT = −νεL (2.37)

where ν = −ε

T

εL

is known as Poisson's ratio.

For the results for three different uniaxial stresses applied separately (see Fig.

2.17) in each of the orthogonal direction, x, y and z can give the general relations

between normal stress and normal strain know as generalized Hooke's law if the strain

components for each of the stress conditions are superposed:

εx = 1E

σx−ν σ

y+σ

z( )( )εy = 1

Eσ y −ν σ x +σ z( )( )

εz = 1E

σ z −ν σ x +σ y( )( )(2.38)

Since shear stresses are decoupled (i.e., unaffected) by stresses in other directions

the relations for shear stress-shear strain are:

γ xy = 1G

τxy

γ xz = 1G

τxz

γ yz = 1G

τyz

(2.39)

where the shear modulus is G =∆τ∆γ =

E

2(1+ ν).

2.18

:

X

Y

Z

σ

ε

ε

εε

ε

ε

εε

ε

σ

σ

y

y y

y x

x

x

x

z

z

z

zStress εx εy εz

σ x Eσ x - νεx = -νEσx - νεx = -νEσ x

σ y -νεy = -νEσy Eσy -νεy = -νEσy

σz - νεz = -νEσ z -νεz = -νEσ z Eσz

Figure 2.17 Development of generalized Hooke's law.

Generalized Hooke's law is usually written in matrix form such that:ε{ } = S[ ] σ{ } (2.40)

where ε{ } is the strain vector, σ{ } is the stress vector and S[ ] is the compliance matrix.

Expanded, Eq. 2.40 is:

2.19

εx

εy

εz

γ xy

γ xz

γ yz

=

1E

−νE

−νE

0 0 0

−νE

1E

−νE

0 0 0

−νE

−νE

1E

0 0 0

0 0 0 1G

0 0

0 0 0 0 1G

0

0 0 0 0 0 1G

σx

σy

σz

τxy

τ xz

τ yz

(2.41)

Although Eq. 2.41 is a very convenient form, it is not that useful since stress is

rarely known with strain being the unknown. Instead, strain is usually measured

experimentally and the associated stress needs to be determined. Therefore, the inverse

of the compliance matrix needs to be found such thatσ{ } = C[ ] ε{ } (2.42)

where C[ ] is the stiffness matrix such that C[ ] = S[ ]−1. Expanding Eq. 2.42 gives

σx

σy

σz

τ xy

τ xz

τ yz

=

E(1+ν)

1+ ν(1−2ν )

νE(1+ν )(1−2ν )

νE(1+ν )(1−2ν )

0 0 0

νE(1+ν )(1−2ν)

E(1+ν )

1+ ν(1−2ν )

νE(1+ν )(1−2ν )

0 0 0

νE(1+ν )(1−2ν)

νE(1+ν )(1−2ν )

E(1+ν )

1+ ν(1−2ν )

0 0 0

0 0 0 G 0 0

0 0 0 0 G 0

0 0 0 0 0 G

εx

εy

εz

γ xy

γ xz

γ yz

(2.43)

Generalized Hooke's can be simplified somewhat for the special case of planestress in the x-y plane since σz =0. Being orthogonal to the x-y plane, Since σz is a also

a principal stress by definition, all the shear stresses associated with the z-direction are

also zero. Thus, the stress-strain relations for plane stress in the x-y plane become

σx

σy

τxy

=

E

(1− ν2 )

νE

(1− ν2)0

νE

(1− ν2)

E

(1− ν2)0

0 0 G

εx

εy

γ xy

(2.44)

For the special case of plane stress, although σz =0, the strain in the z-direction is

not zero but instead can be determined such that

Plane stress : σ z = 0, εz ≠ 0 =−ν

1− ν(εx + εy) (2.45)

2.20

Similarly for the special case of plane strain, although, εz =0, the stress in the z-

direction is not zero but instead can be determined such that

Plane strain : εz = 0, σ z ≠ 0 = ν(σx +σ y) (2.46)

For the special case of hydrostatic pressure, no distortion takes place, only size

changes. In this case the hydrostatic stress is

σH =σ x +σ y + σx( )

3(2.47)

and the dilatation (i.e., volumetric change) is

εV =∆VV

= εx + εy + εx( ) (2.48)

The ratio between the hydrostatic stress and the dilation is a special combination of

elastic constants called the bulk modulus.

k =σH

εV=

σ x + σy + σx( )3 εx + εy + εx( ) =

E

3(1− 2ν )(2.49)

It is useful to know that elastic constants are related to the atomic structure of the

material and thus are not affected by processing or component fabrication. For example,

E is related to the repulsion/attraction between two atoms. The force-displacement curve

for this interaction is shown in Fig. 2.18. Since this is the uniaxial case, recall that

σ =PA

and ε =∆L

L=

x

xe. Therefore, since E is defined as E =

∆σ∆ε =

dσdε with

dσ =dPA

and ε =dx

xe then E =

dσdε =

xe

A

dP

dx. Note from Fig. 2.18 that

dP

dx is the slope of

the force-displacement curve, thus making the elastic modulus E fixed by atomic

interaction. From a materials standpoint, covalent and ionic bonds such as those in

ceramics are stiff leading to high elastic moduli in those materials. Metallic bonds are

intermediate in stiffness leading to intermediate elastic moduli in metals. Secondary

bonds such as those found in polymers are least stiff leading to low elastic moduli in those

materials.

For Poisson's ratio, it is useful to consider the sphere model of atomic structure as

shown in Fig. 2.19. Before deformation the triangle between centers has a transverse

side length of 3R , a longitudinal side length of R and a hypotenuse of 2R. After

longitudinal deformation, the transverse side length is 3R -dx, the longitudinal side

2.21

Figure 2.18 Repulsion-attraction force-displacement curve between two atoms.

length is R+dy and the hypotenuse is still 2R. Applying Pythagorean's theorem after

deformation gives

2R( )2 = 3R − dx( )2 + R + dy( )2 (2.50)

Expanding Eq. 2.50 and eliminating higher order terms leaves

2Rdy = 2 3Rdx ⇒dy

dx= 3 . (2.51)

Recalling that the longitudinal and transverse strains can be written in terms of the

deformations

εy =R + dy( ) − R

R=

dy

R⇒ dy = Rεy

εx =3R − dx( ) − 3R

3R= −dx

3R⇒ dx = − 3Rεx

(2.52)

Combining the two terms for dy and dx in Eq. 2.52 and equating them to Eq. 2.51 gives

dy

dx=

Rεy

− 3Rεx

= 3 ⇒ −εy

εx

≡ −1ν

= 3 (2.53)

which shows that from a simple sphere model and expected deformations, Poisson's ratio

is fundamentally linked to atomic interactions giving a value of 1/3 which is the range of

many dense materials.

2.22

Figure 2.19 Undeformed and deformed sphere model for atomic structure anddetermination of Poisson's ratio

2.23